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The number of ways of choosing with parameters


Different ways of coloring a $4 times 4$ game boardWhat does the “n choose multiple numbers” symbol stands for?Recurrence Relation, Discrete Math problem(Homework)Condition over CombinationsWith how many different ways can Adriana be dressed…????Counting the number of trials.In how many ways we can color $15$ eggs..In how many ways can two different colored balls be chosen?Combinatorics: Coloring a prismProbability of picking two marbles each from two colors when selecting $4$ marbles out of $30$ marbles













4












$begingroup$



At our disposal is a collection of $10$ red, $11$ blue and $12$ yellow fabrics. (each fabric is unique) In how many ways can we choose $4$ different fabrics if we want at least one fabric of each of the three colors?




My solution was since the first fabric chosen must be red, there are $10$ options for it. Then the next fabric must be blue, which has $11$ options. The third fabric is yellow, with $12$ options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are $(9+10+11-3)= 30$ ways to choose the last one, making the total number of choices $9cdot 10cdot 11cdot 30$.



My professor said that I needed to divide that by $2$ to get the right answer, but I just don't understand why. Any help would be much appreciated!










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    4












    $begingroup$



    At our disposal is a collection of $10$ red, $11$ blue and $12$ yellow fabrics. (each fabric is unique) In how many ways can we choose $4$ different fabrics if we want at least one fabric of each of the three colors?




    My solution was since the first fabric chosen must be red, there are $10$ options for it. Then the next fabric must be blue, which has $11$ options. The third fabric is yellow, with $12$ options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are $(9+10+11-3)= 30$ ways to choose the last one, making the total number of choices $9cdot 10cdot 11cdot 30$.



    My professor said that I needed to divide that by $2$ to get the right answer, but I just don't understand why. Any help would be much appreciated!










    share|cite|improve this question









    New contributor




    cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4





      $begingroup$



      At our disposal is a collection of $10$ red, $11$ blue and $12$ yellow fabrics. (each fabric is unique) In how many ways can we choose $4$ different fabrics if we want at least one fabric of each of the three colors?




      My solution was since the first fabric chosen must be red, there are $10$ options for it. Then the next fabric must be blue, which has $11$ options. The third fabric is yellow, with $12$ options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are $(9+10+11-3)= 30$ ways to choose the last one, making the total number of choices $9cdot 10cdot 11cdot 30$.



      My professor said that I needed to divide that by $2$ to get the right answer, but I just don't understand why. Any help would be much appreciated!










      share|cite|improve this question









      New contributor




      cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      At our disposal is a collection of $10$ red, $11$ blue and $12$ yellow fabrics. (each fabric is unique) In how many ways can we choose $4$ different fabrics if we want at least one fabric of each of the three colors?




      My solution was since the first fabric chosen must be red, there are $10$ options for it. Then the next fabric must be blue, which has $11$ options. The third fabric is yellow, with $12$ options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are $(9+10+11-3)= 30$ ways to choose the last one, making the total number of choices $9cdot 10cdot 11cdot 30$.



      My professor said that I needed to divide that by $2$ to get the right answer, but I just don't understand why. Any help would be much appreciated!







      combinatorics discrete-mathematics






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      edited 1 hour ago









      Vinyl_coat_jawa

      3,0101132




      3,0101132






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      asked 10 hours ago









      cmplxlizcmplxliz

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          $begingroup$

          Because under your scheme you would count, for example, both
          $$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
          But these are actually the same choice and therefore should not be counted twice.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$






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              5












              $begingroup$

              Because under your scheme you would count, for example, both
              $$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
              But these are actually the same choice and therefore should not be counted twice.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                Because under your scheme you would count, for example, both
                $$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
                But these are actually the same choice and therefore should not be counted twice.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Because under your scheme you would count, for example, both
                  $$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
                  But these are actually the same choice and therefore should not be counted twice.






                  share|cite|improve this answer









                  $endgroup$



                  Because under your scheme you would count, for example, both
                  $$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
                  But these are actually the same choice and therefore should not be counted twice.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 10 hours ago









                  DavidDavid

                  69.2k667130




                  69.2k667130























                      3












                      $begingroup$

                      The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$






                          share|cite|improve this answer









                          $endgroup$



                          The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 10 hours ago









                          Dbchatto67Dbchatto67

                          1,142118




                          1,142118






















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